But where's the orthocenter?

$AB=\sqrt{(3-1)^2+(6-2)^2}=\sqrt{2^2+4^2}=2 \sqrt{5} \\ AC=\sqrt{(5-1)^2+(4-2)^2}=\sqrt{4^2+2^2}=2 \sqrt{5} \\ \implies AB=AC$

It therefore follows that $\triangle ABC$ is isosceles. Consider the line connecting $A$ to the midpoint of $BC$ .

By definition, the centroid is on this this line. Also, because the triangle is isosceles, the line is perpedicular to $BC$ so the orthocentre is on this line.

The slope of $BC$ is $\dfrac{4-6}{5-3}=\dfrac{-2}{2}=-1$ . The slope of the normal to this line is $-\dfrac{1}{-1}=1$ . So the answer is:

$\color{#20A900}{\boxed{\boxed{1}}}$

Coordinates of centroid is $(1+3+5)/3,(2+4+6)/3=(3,4)$ .

Here are the steps to find the coordinates of circumcenter(copied from a website as I am to lazy to type):

Step 1 : From the coordinates we have to find the slopes and midpoints of the lines.

Step 2 : Find the slope of the bisectors and now find the equations of two lines by using slope and one mid points.

Step 3 : Solve any two pair of equations and find the intersection point.

Step 4 : The point thus found is the circumcenter of the given triangle.

We find the coordinates of the circumcenter is $8/3, 11/3$ .

Slope is $\dfrac{4-11/3}{3-8/3}=1$ .